The shape of the lesson
By the end of the lesson, learners can
- Displacement, velocity, acceleration in step: Set an oscillation going and three curves emerge together.
- The equations behind the curves: With angular frequency ω = 2π/T = 2πf, the displacement is x = x₀ sin ωt (taking x = 0 at t = 0).
- Defining condition: a = −ω²x, with ω = 2πf = 2π / T.
- Velocity: v = ±ω√(x₀² − x²), greatest (ωx₀) at the centre and zero at the extremes.
- Energy: total E = ½mω²x₀², shared between KE (greatest at the centre) and PE (greatest at the extremes).
One picture every idea returns to
Simple harmonic motion is any oscillation in which the acceleration is proportional to the displacement and directed toward equilibrium: a = −ω²x. The solution is x = x₀ sin ωt, with v = ±ω√(x₀² − x²).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, The defining condition.. One condition, a = −ω²x, captures every simple harmonic oscillation. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Displacement, velocity, acceleration in step | Set an oscillation going and three curves emerge together. Project tab 1 of the lab (Simple harmonic motion) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: The equations behind the curves | With angular frequency ω = 2π/T = 2πf, the displacement is x = x₀ sin ωt (taking x = 0 at t = 0). | Connect-Extend-Challenge |
| 30 to 43 min | Check and plenary | The quick-check questions from the student lesson, answered on mini whiteboards one at a time; re-teach the one most get wrong, then a random call on the last. | Individual, then whole class |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 67 marks) from the topic practice set. Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in The equations behind the curves, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Connect-Extend-Challenge
After the new idea, each student writes three lines: how it connects to what they knew, how it extends their thinking, and what still challenges them. Pairs compare, then the challenges become the plenary.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
No examiner trap is recorded for this lesson. Use the quiz key below to spot the distractor most of the class picks.
Simple harmonic motion: topic worksheet
6 questions, 67 marks. Written in the style of the papers; no past paper question is reproduced. These questions cover the whole topic, so sibling lessons share them.
A mass on a spring oscillates with simple harmonic motion of amplitude x₀ = 4.0 cm and period T = 0.80 s. Fig. 1.1 shows how its displacement varies with time.
(a) State the defining equation of simple harmonic motion and explain what each symbol represents. [2]
(b) Calculate the angular frequency ω. [1]
(c) Calculate the maximum acceleration of the mass. [2]
(d) Calculate the maximum speed of the mass. [2]
(e) Calculate the speed of the mass when its displacement is 2.0 cm. [3]
(f) State the displacement at which the speed is greatest, and the displacement at which the acceleration is greatest. [2]
(g) State the phase relationship between the acceleration and the displacement. [1]
A body performs simple harmonic motion with frequency 2.0 Hz and amplitude 5.0 cm.
(a) State the defining equation of SHM and explain the meaning of the minus sign. [2]
(b) State the two conditions required for SHM. [2]
(c) Calculate the angular frequency. [2]
(d) Calculate the maximum acceleration. [2]
(e) Calculate the maximum speed. [2]
A mass oscillates in SHM with amplitude 8.0 cm and period 0.50 s. At t = 0 it is at maximum displacement.
(a) Calculate the angular frequency. [2]
(b) Write the equation for the displacement x as a function of time. [2]
(c) Calculate the displacement at t = 0.10 s. [2]
(d) Calculate the speed when the displacement is 4.0 cm. [2]
(e) State where the speed is maximum and where the acceleration is maximum. [2]
A 0.20 kg mass performs SHM with amplitude 6.0 cm and frequency 5.0 Hz. Parts (a) and (b) use the kinematics; parts (c) to (e) use the energy.
(a) Calculate the angular frequency. [1]
(b) Calculate the speed when the displacement is 3.0 cm. [3]
(c) Calculate the kinetic energy at this displacement. [2]
(d) Calculate the total energy of the oscillation. [2]
(e) Calculate the potential energy at x = 3.0 cm and confirm that KE + PE equals the total energy. [3]
A 0.50 kg trolley on springs oscillates in SHM, completing 20 oscillations in 16 s with amplitude 4.0 cm. Part (a) is kinematics; the rest use the energy.
(a) Calculate the period and the angular frequency. [2]
(b) Calculate the maximum speed. [2]
(c) Calculate the maximum kinetic energy (the total energy). [2]
(d) Calculate the displacement at which the kinetic energy equals the potential energy. [2]
(e) Calculate the kinetic energy when the displacement is 2.0 cm. [3]
A mass on a spring is driven by an external periodic force of variable frequency. Fig. 10.1 shows the response. This question links the SHM, its energy and resonance.
(a) State what is meant by the natural frequency of the system. [1]
(b) The mass is 0.25 kg and the spring gives ω = 20 rad s⁻¹. Calculate the natural frequency. [2]
(c) Explain what is meant by resonance and state the driving frequency at which it occurs. [2]
(d) At resonance the amplitude is 6.0 cm. Calculate the total energy of the oscillation. [3]
(e) Describe and explain how increasing the damping changes the resonance peak, as in Fig. 10.1. [2]
(f) State one situation where resonance is useful and one where it must be avoided. [2]
Answer key and marking notes
- (a)a = −ω²x ✓: the acceleration a is proportional to the displacement x from equilibrium and directed toward equilibrium, with ω the angular frequency ✓
- (b)ω = 2π / T = 2π / 0.80 = 7.9 rad s⁻¹ ✓
- (c)a_max = ω²x₀ = 7.85² × 0.040 ✓ = 2.5 m s⁻² ✓
- (d)v_max = ωx₀ = 7.85 × 0.040 ✓ = 0.31 m s⁻¹ ✓
- (e)v = ω√(x₀² − x²) = 7.85 × √(0.040² − 0.020²) ✓ = 7.85 × √(1.2 × 10⁻³) ✓ = 0.27 m s⁻¹ ✓
- (f)Speed is greatest at x = 0 (the centre) ✓; acceleration is greatest at x = ±x₀ (the extremes) ✓
- (g)They are exactly out of phase (antiphase, 180° apart): a = −ω²x ✓
- (a)a = −ω²x ✓; the minus sign shows the acceleration is always directed opposite to the displacement, toward equilibrium ✓
- (b)The acceleration is proportional to the displacement ✓, and is always directed toward a fixed equilibrium point ✓
- (c)ω = 2πf = 2π × 2.0 ✓ = 13 rad s⁻¹ ✓
- (d)a_max = ω²x₀ = 12.57² × 0.050 ✓ = 7.9 m s⁻² ✓
- (e)v_max = ωx₀ = 12.57 × 0.050 ✓ = 0.63 m s⁻¹ ✓
- (a)ω = 2π/T = 2π/0.50 ✓ = 13 rad s⁻¹ ✓
- (b)x = x₀ cos(ωt) = 0.080 cos(12.6 t) ✓✓
- (c)x = 0.080 cos(12.57 × 0.10) = 0.080 cos(1.257 rad) ✓ = 0.025 m (2.5 cm) ✓
- (d)v = ω√(x₀² − x²) = 12.57 × √(0.080² − 0.040²) ✓ = 0.87 m s⁻¹ ✓
- (e)Speed is maximum at the equilibrium position (x = 0) ✓; acceleration is maximum at the extremes (x = ±x₀) ✓
- (a)ω = 2πf = 2π × 5.0 = 31.4 rad s⁻¹ ✓
- (b)v = ω√(x₀² − x²) = 31.4 × √(0.060² − 0.030²) ✓✓ = 1.6 m s⁻¹ ✓
- (c)KE = ½mv² = ½ × 0.20 × 1.63² ✓ = 0.27 J ✓
- (d)E = ½mω²x₀² = ½ × 0.20 × 31.4² × 0.060² ✓ = 0.36 J ✓
- (e)PE = ½mω²x² = ½ × 0.20 × 986 × 0.030² = 0.089 J ✓; KE + PE = 0.27 + 0.089 = 0.36 J = E ✓✓
- (a)T = 16/20 = 0.80 s ✓; ω = 2π/0.80 = 7.9 rad s⁻¹ ✓
- (b)v_max = ωx₀ = 7.85 × 0.040 ✓ = 0.31 m s⁻¹ ✓
- (c)KE_max = ½mv_max² = ½ × 0.50 × 0.314² ✓ = 0.025 J ✓
- (d)x = x₀/√2 = 0.040/1.414 ✓ = 0.028 m (2.8 cm) ✓
- (e)KE = ½mω²(x₀² − x²) = 0.25 × 61.6 × (0.040² − 0.020²) ✓✓ = 0.018 J ✓
- (a)The frequency at which the system oscillates freely when displaced and released ✓
- (b)f₀ = ω/2π = 20/(2π) ✓ = 3.2 Hz ✓
- (c)The amplitude becomes very large when the driving frequency equals the natural frequency ✓, when energy transfer to the oscillator is greatest; this occurs at f₀ ✓
- (d)E = ½mω²x₀² = ½ × 0.25 × 20² × 0.060² ✓✓ = 0.18 J ✓
- (e)The peak becomes lower and broader ✓, with the maximum amplitude reduced (and shifted to a slightly lower frequency) ✓
- (f)Useful: tuning a radio or a musical instrument; to be avoided: bridges or buildings driven by wind or earthquakes ✓✓
The lesson's own checks as a quiz
The quick-check and exam-style questions from the student lesson, printable with the key.
Simple harmonic motion is defined by the condition:
- Aa = −ω²x
- Ba = ω²x
- Ca = −ωx
- Dv = −ω²x
The angular frequency ω of an oscillation of period T is:
- A2π / T
- B1 / T
- C2πT
- Dπ / T
An object in SHM has its maximum speed when its displacement is:
- Azero, at the centre
- Bequal to the amplitude
- Chalf the amplitude
- Dat either extreme
Using v = ±ω√(x₀² − x²), the speed at the amplitude (x = x₀) is:
- Azero
- Bωx₀
- Cω²x₀
- Dmaximum
Quiz key
The defining condition is a = −ω²x: acceleration proportional to displacement and directed back toward equilibrium.
ω = 2π / T = 2πf, the rate of change of phase in radians per second.
The speed v = ω√(x₀² − x²) is greatest at x = 0 and zero at the extremes.
At x = x₀ the bracket is zero, so v = 0: the object is momentarily at rest at the extreme of its motion.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.